Question

Algorithms
prove that log(n(factorial)) = theta of (nlogn) is true, using Stirling's approximation, which is n(factorial)=(sqrt(2pi*n)) * ((n/e)^n) * (1 + theta of (1/n))

Answer 1

algorithm question*

Answer 2

\[\log n!=\Theta(n\log{n})\]
Is this what you mean?

Answer 3

And Stirling's approximation is
\[n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \left(1+\Theta\left(\frac{1}{n}\right)\right)\]

Answer 4

yes